ASSUMING that "the hash function is fair & uniformly distributes the universe of keys to all the 10 locations",
the probability should be $6/10$.
Due to uniform distribution, the probability that the new record will be placed to any location $i$ will be same for any $i \ \epsilon \ (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)$ & will be equal to $\frac{1}{10}$.
Now if the hash function will try to place the new record into any of the locations $1, 7, 8, 9, 10$, the record will be redirected to location $2$ due to collision resolution by linear probing.
And if the hash function will try to place the new record into $2$, since location $2$ is available, the record will go to location $2$.
So The probability of going new record into location $2$, will be same as probability that the output of hash function is any one of $(1, 2, 7, 8, 9, 10)$ out of $(1, 2, 3, 4, 5, 6, 7, ,8 ,9, 10)$, where each location is equally likely.
So the chances will be $\frac{6}{10}$.